Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-9x-3y &= -5 \\ -7x-y &= 5\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-7x = y+5$ Divide both sides by $-7$ to isolate $x$ $x = {-\dfrac{1}{7}y - \dfrac{5}{7}}$ Substitute this expression for $x$ in the first equation. $-9({-\dfrac{1}{7}y - \dfrac{5}{7}}) - 3y = -5$ $\dfrac{9}{7}y + \dfrac{45}{7} - 3y = -5$ Simplify by combining terms, then solve for $y$ $-\dfrac{12}{7}y + \dfrac{45}{7} = -5$ $-\dfrac{12}{7}y = -\dfrac{80}{7}$ $y = \dfrac{20}{3}$ Substitute $\dfrac{20}{3}$ for $y$ in the top equation. $-9x-3( \dfrac{20}{3}) = -5$ $-9x-20 = -5$ $-9x = 15$ $x = -\dfrac{5}{3}$ The solution is $\enspace x = -\dfrac{5}{3}, \enspace y = \dfrac{20}{3}$.